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3 years ago

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A married couple agreed to continue bearing a new child until they get two boys , but not more than 4 children. Assuming that ea

ch time that a child is born, the probability that is a boy is 0.5 independent from all other times. Find the probability that the couple has atleast two girls. A. 1/2 B. 5/16 C. 5/8 D. 4/15

2 answers:

zlopas [31]3 years ago

6 0

A is the one I think of

Whitepunk [10]3 years ago

6 0

Answer:

A. 1/2 = 0.5

Step-by-step explanation:

We are given that the couple stops bearing a child until they get two boys.

Now, if the couple bears less than 2 girls.

Then the possible cases are: BB, BGB and GBB.

As the probability of the child being a boy = 0.5

Thus, the probability of the child being a girl = 1 - 0.5 = 0.5.

Hence, the probability that the couple bears less than 2 girls = sum of probabilities of BB, BGB and GBB.

i.e. Probability of less than 2 girls = $(\frac{1}{2})^{2}$ + $(\frac{1}{2})^{3}$ + $(\frac{1}{2})^{3}$

i.e. Probability = $\frac{1}{4} +\frac{1}{8} +\frac{1}{8}$

i.e. Probability = $\frac{1}{4} +\frac{2}{8}$

i.e. Probability = $\frac{1}{4} +\frac{1}{4}$

i.e. Probability = $\frac{2}{4}$

i.e. Probability = $\frac{1}{2}= 0.5$

Thus, the probability of less than 2 girls = 0.5

So, the probability of atleast 2 girls = 1 - Probability of less than 2 girls

i.e. The probability of atleast 2 girls = 1 - 0.5 = 0.5

Hence, the probability that the couple has atleast 2 girls is 0.5 = 1/2.

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(a): Marginal pmf of x

$P(0) = 0.72$

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(b): Marginal pmf of y

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Step-by-step explanation:

Given

<em>x = bottles</em>

<em>y = carton</em>

<em>See attachment for complete question</em>

<em />

Solving (a): Marginal pmf of x

This is calculated as:

$P(x) = \sum\limits^{}_y\ P(x,y)$

So:

$P(0) = P(0,0) + P(0,1)$

$P(0) = 0.63 + 0.09$

$P(0) = 0.72$

$P(1) = P(1,0) + P(1,1)$

$P(1) = 0.18 + 0.10$

$P(1) = 0.28$

Solving (b): Marginal pmf of y

This is calculated as:

$P(y) = \sum\limits^{}_x\ P(x,y)$

So:

$P(0) = P(0,0) + P(1,0)$

$P(0) = 0.63 + 0.18$

$P(0) = 0.81$

$P(1) = P(0,1) + P(1,1)$

$P(1) = 0.09 + 0.10$

$P(1) = 0.19$

Solving (c): Mean and Variance of x

Mean is calculated as:

$E(x) = \sum( x * P(x))$

So, we have:

$E(x) = 0 * P(0) + 1 * P(1)$

$E(x) = 0 * 0.72 + 1 * 0.28$

$E(x) = 0 + 0.28$

$E(x) = 0.28$

Variance is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Calculate $E(x^2)$

$E(x^2) = \sum( x^2 * P(x))$

$E(x^2) = 0^2 * 0.72 + 1^2 * 0.28$

$E(x^2) = 0 + 0.28$

$E(x^2) = 0.28$

So:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = 0.28 - 0.28^2$

$Var(x) = 0.28 - 0.0784$

$Var(x) = 0.2016$

Solving (d): Mean and Variance of y

Mean is calculated as:

$E(y) = \sum(y * P(y))$

So, we have:

$E(y) = 0 * P(0) + 1 * P(1)$

$E(y) = 0 * 0.81 + 1 * 0.19$

$E(y) = 0+0.19$

$E(y) = 0.19$

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$Var(y) = E(y^2) - (E(y))^2$

Calculate $E(y^2)$

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$E(y^2) = 0^2 * 0.81 + 1^2 * 0.19$

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$E(y^2) = 0.19$

So:

$Var(y) = E(y^2) - (E(y))^2$

$Var(y) = 0.19 - 0.19^2$

$Var(y) = 0.19 - 0.0361$

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Solving (e): The covariance and the coefficient of correlation

Covariance is calculated as:

$COV(x,y) = E(xy) - E(x) * E(y)$

Calculate E(xy)

$E(xy) = \sum (xy * P(xy))$

This gives:

$E(xy) = x_0y_0 * P(0,0) + x_1y_0 * P(1,0) +x_0y_1 * P(0,1) + x_1y_1 * P(1,1)$

$E(xy) = 0*0 * 0.63 + 1*0 * 0.18 +0*1 * 0.09 + 1*1 * 0.1$

$E(xy) = 0+0+0 + 0.1$

$E(xy) = 0.1$

So:

$COV(x,y) = E(xy) - E(x) * E(y)$

$Cov(x,y) = 0.1 - 0.28 * 0.19$

$Cov(x,y) = 0.1 - 0.0532$

$Cov(x,y) = 0.0468$

The coefficient of correlation is then calculated as:

$r = \frac{Cov(x,y)}{\sqrt{Var(x) * Var(y)}}$

$r = \frac{0.0468}{\sqrt{0.2016 * 0.1539}}$

$r = \frac{0.0468}{\sqrt{0.03102624}}$

$r = \frac{0.0468}{0.17614266944}$

$r = 0.26569371378$

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